Problem: Suppose $G$ is a group, and $H_1, H_2$ are subgroups of $G$. Prove that if $H_1, H_2$ are normal subgroups of $G$, $H_1H_2=\{xy: x\in H_1, y\in H_2\}=G$, and $H_1\cap H_2=\{1\}$ where $1$ is the identity, then $G/H_1\simeq H_2$.
My Question: If we define $f: G/H_1 \to H_2$ to be $f(\sigma H)=\sigma$ where $\sigma\in G$, then $f$ might not be well-defined. Is there a theorem that could be of significant help here? Any pointer would be greatly appreciated.
If you define the mapping in opposite direction, it is going to be easier and clearer: $$ g:H_2 \to G/H_1, \quad g(y)=y H_1. $$ Obviously, this is an homomorphism.
If $x \in H_1, \ y \in H_2$, then $xy=yx$. To show this, we write $yx=uv$, where $u \in H_1, \ v \in H_2$. By normality, $x^{-1}uv=x^{-1}yx \in H_2$, thus, $x^{-1}u \in H_2 \ \Rightarrow x^{-1}u=1 \ \Rightarrow x=u$. Similarly, $y=v$.
Surjectivity: Let $zH_1 \in G/H_1$. Then $z=xy=yx$ where $x\in H_1, \ y\in H_2$. Thus, $zH_1 = yxH_1 = yH_1 = g(y)$.
Injectivity: If $g(y)=g(u)$, then $yH_1=uH_1$ and this implies $u^{-1}y\in H_1$. Hence, $u^{-1}y=1 \Rightarrow y=u$.
Your question is answered now because $f=g^{-1}$.